Path Integrals
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Let's start off with a video. This is a great video which includes a complete introduction, with examples, and explanation of applications of path integrals. It is well worth your time to watch it.
video by Dr Chris Tisdell |
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The equations from the last video are important for evaluating path integrals. Here is a summary.
requirements |
result | |
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continuous scalar function \(f(x,y,z)\) |
path integral of \(f\) over \( \mathcal{C} \) | |
curve \( \mathcal{C} \) parameterized as \(\vec{c}(t)\) on an interval \(a \leq t \leq b\) |
\(\displaystyle{ \int_{\mathcal{C}}{ f(x,y,z)~ds} = }\) \(\displaystyle{ \int_{a}^{b}{ f(\vec{c}(t)) ~ \| \vec{c}'(t) \| ~dt } }\) | |
continuous \(\vec{c}'(t)\) [derivative of \(\vec{c}(t)\) with respect to \(t\)] |
Most of the above equations should be familiar to you. However, one comment is in order about the term \(f(\vec{c}(t))\). How do you substitute a vector into a function?
If we write the vector as \( \vec{c}(t) = \langle X(t), Y(t), Z(t) \rangle = X(t) \hat{i} + Y(t) \hat{j} + Z(t)\hat{k} \) then we substitute the vector components into \(f(x,y,z)\) as \(x=X(t)\), \(y=Y(t)\) and \(z=Z(t)\). Another way to write this is
\(f(x,y,z) = f(X,Y,Z) = f(X(t), Y(t), Z(t))\). The result is a function of \(t\) only. All \(x\)'s, \(y\)'s and \(z\)'s will be gone, leaving only an integral in \(t\).
A very interesting comment he makes in the above video is that the direction of the curve does not affect the answer, i.e. integrating in the opposite direction will give the same answer. This is true because we use only the magnitude of the derivative of the parameterized curve, i.e. \(\| \vec{c}'(t) \|\), which is the same regardless of the direction. However, the path integral is dependent on the chosen path, i.e. different paths will, in general, give different results. So path integrals are sensitive to the paths.
Okay, let's try some practice problems.
Practice
Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.
Problem Statement
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.
Problem Statement
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.
Problem Statement
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell |
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Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^2-1+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.
Problem Statement
Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^2-1+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\) - \((1,1)\) and \((1,1)\) - \((1,0)\), giving your answer in exact terms.
Problem Statement
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\) - \((1,1)\) and \((1,1)\) - \((1,0)\), giving your answer in exact terms.
Solution
video by Dr Chris Tisdell |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.
Problem Statement
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.
Solution
video by PatrickJMT |
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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Problem Statement |
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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Final Answer |
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\((5\sqrt{5}+11)/6\)
Problem Statement
Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Solution
video by Michel vanBiezen |
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Final Answer
\((5\sqrt{5}+11)/6\)
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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).
Problem Statement |
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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).
Hint |
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There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1-t)x_1+tx_2\)
\(y=(1-t)y_1+ty_2\)
\(z=(1-t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Problem Statement
Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).
Hint
There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1-t)x_1+tx_2\)
\(y=(1-t)y_1+ty_2\)
\(z=(1-t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Solution
video by MIP4U |
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Practice Instructions
Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.